Why is $5$ the remainder of $5 \over 13$?

Question

I don't understand why the remainder of $5 \over 13$ is $5$. I know that the DA tells us that $5 = 0(13) + r$ so the remainder has to be $5$ based on this, but I'm a little unsure of why/how it works

Answer 1

Well $r$ has to be in a set $\{0,1,2...,12\}$. Which of this fullfiles equation you wrote?

And this is always true if you divide $a$ by $b$ and $a<b$. You will get $r=a$:

$$ a = 0\cdot b +r\implies r=a$$

Answer 2

For positive integers, "divide by $13$" means "subtract the largest multiple of $13$ you can while still remaining nonnegative". The remainder is what's left. If you think in terms of objects, you remove them from the set you start with, in batches of $13$.

If you start with $5$ you can't subtract any multiples of $13$ so all $5$ are left.

You don't need a formal "division algorithm" for this. It works fine if all you know about division is repeated subtraction (which is what the division algorithm formalizes).

Answer 3

For an intuitive approach: If you have five candies to give out equally to 13 children, each child gets 0 candies, and there remain 5 undistributed candies.